Regardless of the performance, you should be able to do this using loc with boolean indexing:

df = pd.DataFrame([[5, 2], [3, 4]], columns=('a', 'b'))

# modify value in column b where a is 3
df.loc[df.a == 3, 'b'] = 6

df
#   a   b
#0  5   2
#1  3   6

.loc is an indexer. It looks for an entry in the index, but the column name is not an index. It is simply a column. The following solutions would work:

df.loc[4, 'rating'] = 100 # Because 4 is in the index, but how do you know?

or:

df.loc[df['name']=='cheerio', 'rating'] = 100 # Find the row by column

or:

df.set_index('name', inplace=True) # Make 'name' the index
df.loc['cheerios', 'rating'] = 100     # Use the indexer

Try this:-

Solution1:-

pandas.concat([df['type'].dropna(), df['caught'].dropna(),df['bowled'].dropna(),df['hit'].dropna()]).reindex_like(df)

If you want that data to become the new column bar, just assign the result to df['dismissal_kind'].

Solution2:-

dataframe["dismissal_kind"] = dataframe["type"].map(str) + dataframe["cought"].map(str)+dataframe["bowled"].map(str)+dataframe["hit"].map(str)